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Suppose I have two std::vectors x and y and a binary function F. I would like to create a new object z (not necessarily a vector), with the property that the ith elemenent of z will be the application of F to the ith elements of x and y. For the time being I use 

boost::range::transform(x, y, z.begin(), F)

which of course requires memory allocation for z. My goal however is to avoid this memory allocation and have the transformation lazy evaluated, similarly to what boost::adaptors::transformed does with unary functions. In other words, the type of z would be some type of boost::range. Is there any easy way to do this or would I have to write my own range class? Thanks in advance!

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Is it easier to use and std::transform with back_inserter to make std::vector z and then make a range from it? –  maverik Jun 7 '13 at 11:07
    
@maverik: thanks for your answer. Unfortunately, that will not achieve my goal since memory would still need to be allocated for z (this time sequentially). –  linuxfever Jun 7 '13 at 11:26
    
Even if one could achieve lazy evaluation, how do you imagine avoiding memory allocation when the vector size is unknown at compile time? I can only think of reserving enough memory once instead of sequentially; and depending on the context, one could leverage alloca(). Are those two possibilities an option? –  klaus triendl Oct 30 '13 at 21:14

1 Answer 1

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The problem is that Boost.Range does not provide a zip_range that can zip x and y as the inputfor your transform. So you need to write one yourself, e.g. using make_iterator_range and zip_iterator (from the Boost.Iterator library). See e.g. this answer for a code example.

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